Palindromic Sums of Squares

A palindromic coincidence occurred with Five Consecutive Integers or is there a pattern ?

99 + 100 + 101 + 102 + 103 = 505
99 ² + 100 ² + 101 ² + 102 ² + 103 ² = 51015

10099 + 10100 + 10101 + 10102 + 10103 = 50505
10099 ² + 10100 ² + 10101 ² + 10102 ² + 10103 ² = 510151015

11 ² + 12 ² + 13 ² = 434
119811 ² + 119812 ² + 119813 ² = 43064746034

37 ² + 38 ² + 39 ² = 4334
36997 ² + 36998 ² + 36999 ² = 4106556014

Messages

[ August 1, 2005 ]
Jean-Claude Rosa (email) ! [ goto entry ]

Cher Patrick,

Voici les numéros 38 et 39 qui hélas ne sont pas premiers
(le numéro 38 a 5 facteurs premiers mais le numéro 39 est
un semi-prime -un "vrai"- avec un facteur de 15 chiffres et
un autre de 17 chiffres ):

numéro 38 :
1682059335368470748635339502861 = 29 * 109 * 35935429 * 206697457 * 71640539617
917076696729469^2 + 917076696729470^2

numéro 39:
3167016920841776771480296107613 = 113665666219493 * 27862564186498841
1258375325735882^2 + 1258375325735883^2

Voilà c'est tout pour les palindromes de la forme 1... ...1 ou
3... ...3 de 31 chiffres. J'ai commencé à chercher ceux de la
forme 5... ...5 .
Pour espèrer trouver un palprime il faut donc s'attaquer aux
nombres de 33 chiffres ( vraiment dur, dur ! )

A bientôt.
JCR

[ September 6, 2005 ]
Jean-Claude Rosa (email) ! [ goto entry ]

Cher Patrick ,

Voici le numéro 40 :

5450871426137276727316241780545 =
1650889370330016^2 + 1650889370330017^2

C'est le dernier des palindromes de 31 chiffres !
J'ai donc commencé la recherche des palindromes de
33 chiffres mais cela risque de prendre pas mal de temps...

A bientôt.

Kind regards.
JCR

[ September 15, 2006 ]
Jean-Claude Rosa (email) ! [ goto entry ]

Cher Patrick,
Ca y est !! J'ai enfin trouvé le numéro 41 le voici :

316370934175751979157571439073613
=12577180410882082^2 + 12577180410882083^2

mais hélas il n'est pas premier (4 facteurs ) et comme j'ai fini l'étude
des palindromes de la forme 31...13 pour trouver la cinquième
solution du puzzle 14 il faudra étudier les nombres de 35 chiffres...
je ne sais pas si j'aurai le temps et le courage !
Je vais quand même essayer de finir l'étude des palindromes de 33 chiffres
de la forme 5.....5.

JCR

[ October 14, 2006 ]
Jean-Claude Rosa (email) ! [ goto entry ]

Cher Patrick,

... et je viens de trouver le numéro 42.
Le voici:
501263304966749757947669403362105
=15831350305118476^2 + 15831350305118477^2

Voilà, c'est tout pour aujourd'hui...
j'espère arriver un jour au numéro 50 ;-)

JCR

[ May 21, 2007 ]
Jean-Claude Rosa (email) ! [ goto entry ]

Cher Patrick,
Ca y est ! J'ai enfin trouvé le numéro 43.
Le voici :
562318868215014101410512868813265 =
16767809460615511^2 + 16767809460615512^2

Amitiés.
JCR

[ January 12, 2008 ]
Jean-Claude Rosa (email) ! [ goto entry ]

Cher Patrick,
... avec un peu de retard voici le numéro 44 :
588186547187469040964781745681885 =
17149147897016181^2 + 17149147897016182^2

Best regards.
JCR

Sources Revealed

Huen Y.K. from Singapore developed a general generating function for palindromic sums and products
of consecutive integers using concise programcode written for Macsyma 2.2.1.
Global Generating Function For Palindromic Sums and Products of Consecutive Integers.

See also Powers of Consecutives Summing to Palindromes and WONplate 145

Neil Sloane's "Integer Sequences" Encyclopedia can be consulted online : Neil Sloane's Integer Sequences The regular numbers of form n2 + (n+1)2 [sums of two squared consecutives] : %N Centered square numbers: 2n(n-1)+1. under A001844. The subsets in relation to the palindromic and prime numbers of above form : %N n^2 + (n+1)^2 is palindromic. under A027571 %N Palindromes of form n^2 + (n+1)^2. under A027572. %N n^2 + (n+1)^2 is prime. under A027861 %N Primes of form n^2 + (n+1)^2. under A027862. The regular numbers of form n2 + (n+1)2 + (n+2)2 [sums of three squared consecutives] : %N Points on surface of square pyramid: 3*n^2 + 2. under A005918. The subsets in relation to the palindromic and prime numbers of above form : %N n^2 + (n+1)^2 + (n+2)^2 is palindromic. under A027573 %N Palindromes of form n^2 + (n+1)^2 + (n+2)^2. under A027574. %N n^2 + (n+1)^2 + (n+2)^2 is prime. under A027863 %N Primes of form n^2 + (n+1)^2 + (n+2)^2. under A027864. The regular numbers of form n2 + (n+1)2 + (n+2)2 + (n+3)2 [sums of four squared consecutives] : %N n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2. under A027575. The subsets in relation to the palindromic numbers of above form : %N n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 is palindromic. under A027576 %N Palindromes of form n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2. under A027577. The regular numbers of form n2 + (n+1)2 + (n+2)2 + (n+3)2 + (n+4)2 [sums of five squared consecutives] : %N n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2. under A027578. The subsets in relation to the palindromic numbers of above form : %N n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2 is palindromic. under A027579 %N Palindromes of form n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2. under A027580. The regular numbers of form n2 + (n+1)2 + (n+2)2 + (n+3)2 + (n+4)2 + (n+5)2 : %N n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2 + (n+5)^2. under A027865. The subsets in relation to the prime numbers of above form : %N n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2 + (n+5)^2 is prime. under A027866 %N Primes of form n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2 + (n+5)^2. under A027867. Click here to view some of the author's [P. De Geest] entries to the table. Click here to view some entries to the table about palindromes.

A simple substitution of n with (m–1) will proof that n^2 + (n+1)^2 + (n+2)^2 equals 3*n^2 + 2 :

n^2 + (n+1)^2 + (n+2)^2 =

(m–1)^2 + ((m–1)+1)^2 + ((m–1)+2)^2 =

(m^2–2m+1) + (m^2) + (m^2+2m+1) =

m^2 + 1 + m^2 + m^2 + 1 =
3*m^2 + 2 QED

n^2 + (n+1)^2 =

(m–1)^2 + ((m–1)+1)^2 =

m^2–2m+1 + m^2 =

2m^2 – 2m + 1 =
2m(m–1) + 1 QED

The Table

Jean-Claude Rosa - goto entry sum of 3 squares - goto entry sum of 2 squares

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(All rights reserved) - Last modified : January 12, 2008.

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Patrick De Geest - Belgium - Short Bio - Some Pictures
E-mail address : pdg@worldofnumbers.com

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